What is the integral from 1 to 4 of (-3)/((2x+1)^2) ?

The integral from 1 to 4 of (-3)/((2x+1)^2) is -1/3

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integral from 1 to 4 of (-3)/((2x+1)^2)

\int_{1}^{4} \frac{- 3}{( 2 x + 1 ) ^{2}} d x

- \frac{1}{3}

Decimal Notation

- 0.33333 \dots

Solution steps

\int_{1}^{4} \frac{- 3}{( 2 x + 1 ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 3 \cdot \int_{1}^{4} \frac{1}{( 2 x + 1 ) ^{2}} d x

Apply u-substitution

: \int_{3}^{9} \frac{1}{2 u ^{2}} d u

= - 3 \cdot \int_{3}^{9} \frac{1}{2 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 3 \cdot \frac{1}{2} \cdot \int_{3}^{9} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{3}^{9}

= - 3 \cdot \frac{1}{2} [- \frac{1}{u}]_{3}^{9}

Simplify

- 3 \cdot \frac{1}{2} [- \frac{1}{u}]_{3}^{9} : - \frac{3}{2} [- \frac{1}{u}]_{3}^{9}

= - \frac{3}{2} [- \frac{1}{u}]_{3}^{9}

Compute the boundaries:

\frac{2}{9}

= - \frac{3}{2} \cdot \frac{2}{9}

Simplify

= - \frac{1}{3}

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What is the integral from 1 to 4 of (-3)/((2x+1)^2) ?
The integral from 1 to 4 of (-3)/((2x+1)^2) is -1/3